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True [87]
3 years ago
10

What is the y-intercept for lines that have the general form, y = mx

Mathematics
1 answer:
Stolb23 [73]3 years ago
4 0

Answer:

It is 0

Step-by-step explanation:

y = mx indicates that the line has no y intercept and that it passes through the origin (0,0).

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Figure ABCD has verticies A(-4, 1) B(2, 1) C(2, -5) D(-4-3). What was the area of Figure ABCD.
JulijaS [17]

area: 18 units^2

Step-by-step explanation:

the shape is a quadrilateral, with one slanted side, so I separated the shape into a rectangle and triangle, and and calculated their area respectively, then added the products up. please correct me if I'm wrong. hope this helped. :)

8 0
3 years ago
Read 2 more answers
Write a trinomial whose binomial factors contain constant terms that sum to −11 and have a product of 18.
yaroslaw [1]

Answer:

The answer is (x - 9)(x - 2), or x2 - 11 + 18.

Step-by-step explanation:

Well, think about it. What two numbers add to -11 and multiply to get 18? -9 and -2 add to get -11 and multiply to get 18.

8 0
3 years ago
A circle with radius of 3cm sits inside a circle with radius of 11 cm? What is the area of the shaded region?
garri49 [273]

Answer:

3cm = 28.26

11cm = 379.94

3 x 3 x 3.14 = 28.26

11 x 11 x 3.14 = 379.94

hope this helps

8 0
3 years ago
Please help! The problem and answer choices are in the attachment below.
Ray Of Light [21]
4) (log3)14/(log3)4

hope dis helps<span />
4 0
3 years ago
) find a vector parallel to the line of intersection of the planes 5x − y − 6z = 0 and x + y + z = 1.
snow_tiger [21]
The cross product of the normal vectors of two planes result in a vector parallel to the line of intersection of the two planes.

Corresponding normal vectors of the planes are
<5,-1,-6> and <1,1,1>

We calculate the cross product as a determinant of (i,j,k) and the normal products

    i   j   k
   5 -1 -6
   1  1  1

=(-1*1-(-6)*1)i -(5*1-(-6)1)j+(5*1-(-1*1))k
=5i-11j+6k
=<5,-11,6>

Check orthogonality with normal vectors using scalar products
(should equal zero if orthogonal)
<5,-11,6>.<5,-1,-6>=25+11-36=0
<5,-11,6>.<1,1,1>=5-11+6=0

Therefore <5,-11,6> is a vector parallel to the line of intersection of the two given planes.
5 0
3 years ago
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